L and F live 28 mi apart. They ride bikes toward each other.

[wandacmc]

This is my son's 6th grade math problem. He is supposed to make a table and show his work. He has started a table, but we can't seem to come up with an exact answer about time.

L and F live 28 miles apart . They agree to meet between their homes by riding their bikes directly toward each other. If L rides at 12MPH and F at 9MPH and if they both leave at 12:00Noon, what time will they meet?

Thank you for your insight. Any help is appreciated!

 

[Deleted member 4993]

Re: Word problem

This is my son's 6th grade math problem. He is supposed to make a table and show his work. He has started a table, but we can't seem to come up with an exact answer about time.

L and F live 28 miles apart . They agree to meet between their homes by riding their bikes directly toward each other. If L rides at 12MPH and F at 9MPH and if they both leave at 12:00Noon, what time will they meet?

Thank you for your insight. Any help is appreciated!

Every 20 miniutes (1/3 hr) L will ride 4 miles and F will ride 3 miles towards each other.

Time ..................................Distance apart(miles)
12:20                                   28 - 4 - 3 = 21
12:40                                   21 - 4 - 3 = 14

Now continue...

It will be useful if your son communicated with us directly - under your supervision.

 

[stapel]

He has started a table, but we can't seem to come up with an exact answer about time.

But... finding the time is the easy part! :shock:

You're an adult, so you know how to use distance and rate to find time. Since they're heading toward each other, you add to find the total speed. Then you find the time by dividing the distance by that speed (just as you'd find the drive time to visit distant family by dividing the distance by your expected average driving rate). :wink:

(Granted, for this particular exercise, you'll get a fractional answer, but you can simply convert that to "hours and minutes", instead of leaving it as just "hours".)

What on earth is he supposed to do for a "table"...? Please clarify. Thank you!

Eliz.

 

[nycfunction]

This is my son's 6th grade math problem. He is supposed to make a table and show his work. He has started a table, but we can't seem to come up with an exact answer about time.

L and F live 28 miles apart . They agree to meet between their homes by riding their bikes directly toward each other. If L rides at 12MPH and F at 9MPH and if they both leave at 12:00Noon, what time will they meet?

Thank you for your insight. Any help is appreciated!

We can use the distance formula, which is D = rt, where D = distance, r = rate (or speed) and t = time (in hours) NOT minutes.

We have L and F traveling toward each other, which means times equal, total distance given.

....................Time..............Rate...................Distance
L....................x..................12mph................28 miles
F....................x..................9mph..................28 miles

12x + 9x = 28 miles

Solve for x.

21x = 28

Divided both sides by 21 to find x.

x = 28/21

We can reduce this fraction to lowest terms.

What is the BIGGEST number that goes into 28 and 21 WITHOUT a remainder?

How about 7?

28 divided by 7 = 4
21 divided by 7 = 3

So, x = 4/3, which means 1 hour and 75 minutes.

They both left at 12noon.

So, 12 noon PLUS 1 hour + 75 minutes = 1:45pm.

So, at what time did L and F meet? They met at 1:45pm.

Is this clear?

 

[stapel]

We can use the distance formula, which is D = rt, where D = distance, r = rate (or speed) and t = time (in hours) NOT minutes....

Actually, since this was posted to the "Arithmetic" category, we cannot assume any familiarity with variables or solving linear equations.

Eliz.